Imagine

Euler's Formula tells us that $e^{i\pi} = -1$

$\Rightarrow e^{\frac{i\pi}{2}} = \sqrt{-1} = i$

$\Rightarrow i^i = \left(e^{\frac{i\pi}{2}}\right)^i = e^{-\frac{\pi}{2}}$

Technically, there's an infinite number of values for $i^i$ , which should be $e^{\frac{i\pi}{2} + 2n\pi}$ , $n$ being an integer.

EDIT: It appears the problem is now edited to ask for the principal value, so the answer is absolutely correct.

Using what we know about the polar form of complex numbers, we can rewrite ${ i }^{ i }$ as $({ e }^{ i\pi /2 }{ ) }^{ i }$ . Using the multiplicative property of exponents, we receive our answer. ${ i }^{ i }={ e }^{ -\pi /2 }$